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F&&uuu&&ck!, some has really totally screwed the search function in Geotech.
I had to find it on my hard disk, which is a much simpler task!
Cheers,
/\sif
By putting an additional emitter resistor (where the main collector current is draining into), we can greatly reduce the op-amp drive current and adjust the output DC voltage at the output of the op-amp to the half supply voltage level. This gives us a wide output rail range too.
That's a little improvement only. But I like it.
Did I win the great prize (Oscar? Nobel prize? The "Snake Oil" prize?) ???
Anyway.
Cheers,
^sif
The idea of allowing bias current to flow through the op amp is related with forcing the op amp output in constant class A operation, thus reducing THD. If you are not biasing your BJT over, say, 10mA, it will not harm the op amp in any way. Besides, you are sinking that current anyway, so consumption-wise there is no real difference whether you sink it through op amp or a resistor.
The idea of allowing bias current to flow through the op amp is related with forcing the op amp output in constant class A operation, thus reducing THD. If you are not biasing your BJT over, say, 10mA, it will not harm the op amp in any way. Besides, you are sinking that current anyway, so consumption-wise there is no real difference whether you sink it through op amp or a resistor.
Hi Davor,
no, I don't agree with you. For the sake of lowering the total harmonic distortion, it is better to relieve the op-amp driver stage.
Anyway.
My new BJT's should arrive today (I can see it in the online tracking system) and let's hope, there is a good BJT there.
Cheers,
Aziz
Wow!, I can't believe it!
Ready for new records?
1 x BJT MJE 13007 (single BJT only!!!) + NE 5534A, gain resistors 470 Ohm, 2.2 Ohm (approx. x 210): en = 0.22 nV/sqrt(Hz)
I have to change my new goal quickly.
More to come in a few days.
Stay tuned..
^sif,
owner of the "World's Quietest Metal Detector Front-End Ever"(c)(r)(tm) *LOL*
I have found the reason, why the calculated voltage noise density becomes so attractive!
The MJE 13007 must have a large current noise density (which must dominate now). The current noise hasn't been taken into account yet. That's very likely the reason, why I get a so nice voltage noise density, which isn't likely correct. I have to change my measurement method to take both voltage and current noise density into account.
Cheers,
Aziz
I have to take the current noise into account now, which have been totally neglected previously.
Three measurements on the FFT spectrum are required now:
Measuring the voltage noise floor level in dB.
- at short input (0 Ohm resistor noise measurement)
- at open input (input impedance resistor noise measurement)
- at a known resistor R connected to the input (0 < R < input impedance of the amplifier)
A new prize to win:
How to calculate the voltage & current noise density of the amplifier system?
Who wants to win that great prize?
I am going to be...
Cheers,
Aziz
I thought something must be amiss. I get more noise than that by probing a piece of wire dangling from the ceiling!
Yep. When I take the current noise effect into account and approximate the voltage noise density, I come to en = 0.42 nV/sqrt(Hz)
This is indicating the correct approximate value. Still excellent performance with one BJT!
I will improve the measurement method soon and we will see much better & more accurate figures.
But someone has to win that great prize before I can calculate them.
Who's going to find the nice noise formula?
^sif
I'll show you the used parameters to give you an initial idea of how to calculate the figures. One measurement data is supplied.
Solve the formulas for en (voltage noise density) and in (current noise density) and win the great prize. Clue:
Noise calculation:
e = sqrt(en^2 + 4*k*T*Rs + (in*Rs)^2), where
e = Total voltage noise density of the amplifier system
en = Voltage noise density of the amplifier system
in = Current noise density of the amplifier system
Rs = Input source resistance
Note, that I'm using the relation factors f1 and f2 of the measurements (see below the picture).
Good luck in solving the equations.
Cheers,
Aziz
PS: Come on guys!, solve the two equations for me please. Someone has to bite the bullet this time!!!!!!
(Don't look at me!!!! . Where are the gurus out there?)
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